Nets. By a net we mean a collection of elements called cells (usually represented as small geometrical areas) some of which are linked to each other (these links being shown by lines, curved or straight, drawn from cell centre to cell centre) and which cannot be divided into two collections of cells with no cell in one linked to any cell in the other; in other words the net is connected. A net of one cell is a singleton. A net of two cells is a domino. A cell linked to one other cell is an end. Thus a domino has two ends. A cell linked to two other cells is a passage. The number of other cells to which a cell is linked is its degree of linkage. A theorem of Euler (expressed in our terminology) is that: the number of cells of odd degree in a net is even (To prove this note that each link connects two cells, so the sum of the degrees of all the cells is twice the number of links, an even number. Sums and differences of even numbers are even. For a sum of odd numbers to be even there must be an even number of them.)
A net in which each cell is linked to no more than two other cells is a path; either an open path with two ends (also called a chain), or a closed path with no ends (also called a circuit). In terms purely of links there are two types of net with three cells, an open or closed path. The property of a net of being connected means that given any two cells an open path can be found, using only the links of the net, with those cells as its ends. A path, open or closed, that visits every cell of the net is what we mean by a tour.
Boards. By a board we mean a net in which the links have specified lengths. Cells whose links are of the minimum length, which we can take to be the unit of length, are said to be adjacent (and are usually represented by the cells having a common edge). By a chessboard we mean a board in which the unit links form straight lines in two mutually perpendicular directions, which we call ranks, shown horizontally across the screen, and files, shown vertically down the screen. (It is also possible to consider boards in which the cells form a hexagonal lattice, hexboards, or a three-dimensional lattices spaceboards; we may add notes on these at a later date.) The ranks and files of a chessboard need not all contain the same number of cells, nor form unbroken straight lines; the outline of the board may be non-rectangular and there may be holes in the board. Any such board however can be expanded to form a rectangular board by adding cells in the holes and round the edges, to form the minimal containing rectangle. The cells on a chessboard can be specified by an ordered pair of numerical coordinates (x,y) that count the number of unit links from a given cell (0,0) in the minimal containing rectangle. The cell taken as (0,0) is usually the lower left corner cell, so that the coordinates are whole numbers. The use of coordinates to record positions and moves in chess goes back at least to the time of al Adli (c.840), although different sequences of letters were used to label the ranks and the files. To apply mathematical methods it is necessary to make both coordinates numerical.
Moves. By a move we mean the transfer of a token from a cell (x,y) to a cell (x',y'), and we represent the move in length and direction by the ordered pair of numbers (x'–x,y'–y) = (u,v), where u and v can be positive, negative or zero. By the pattern of a move (u,v) we mean the pair of non-negative numbers {r,s} where r = |u|, s = |v| or vice versa, and r < s. A move of pattern {r, s} takes a piece from (x,y) to any of the up to eight cells (x±r,y±s) and (x±s,y±r) that are available within the confines of the board. We allow the possibility that the initial and final cells of the move may be the same in which case we have a null move {0,0}. A move that has one coordinate zero, i.e. of pattern {0,s} is termed orthogonal or rookwise, i.e. horizontal or vertical, along the ranks or files of the lattice. A move with both coordinates of equal magnitude, i.e. of pattern {r,r} is termed diagonal or bishopwise, i.e. when the cells are square, lines of cells at 45° to the horizontal and vertical. A move with both coordinates different and non-zero we call skew. For a skew move on a sufficiently large board all eight moves from a given cell will be distinct, making a wheel formation. If r<s then the four moves (±r, ±s) are termed vertical while (±s, ±r) are horizontal. For diagonal or lateral moves the number reduces to four.
Pieces By a (chess)man we mean a token of distinctive design that can be placed to occupy certain cells of a chessboard and whose design indicates the moves that it is permitted to make, subject to the limitations that may be provided by the edges of the board or the presence or absence of other tokens, or any other conditions applying. By a (chess)piece we mean a chessman whose powers are independent of time, position on the board, or orientation, thus if able to make a move (u,v) of pattern {r,s} from a cell (x,y) then it can make any of the moves (±r, ±s) or (±s, ±r) from (x,y) or any other cell, subject to the existence of the destination cell. Given any pattern or set of patterns of moves then a piece can be defined, able to make just those moves from whichever cell of the board it is on. A simple piece is one having only a single pattern of move {r,s}; more complicated pieces are compound. The freedom (of movement) of a piece can be measured by the fraction of the board to which it has access. Conversely the multiplicity of a piece is the number of pieces of that type needed to patrol all the cells of the board. A piece that is able to reach any square of the board from any other square in a series of moves (a requirement for any piece that makes a tour of the whole board) is a free piece; one that can reach all squares of one colour, but not the other colour, may be termed half-free. The mobility of a piece on a given board is the average number of moves it can make when placed at random on the board. This can be calculated by totalling the moves available to it at all the cells of the board and dividing by the number of cells.
A piece that cannot be blocked by men on intermediate cells we call a leaper. The shortest simple leapers have acquired colourful names, some of them dating from mediaeval times, as follows:
{0,0} = dummy, {0,1} = wazir, {1,1} = fers, {0,2} = dabbaba, {1,2} = knight, {2,2} = alfil, {0,3} = threeleaper, {1,3} = camel, {2,3} = zebra, {3,3} = tripper, {0,4} = fourleaper, {1,4} = giraffe, {2,4} = lancer, {3,4} = antelope, {4,4} = commuter.
The dummy merely occupies a cell. It occurs in some versions of chess which apply the rule that a pawn moved to its final rank has to wait there until a captured piece is available for it to promote to. The name wazir comes from a piece that appeared in a variant of chess played on an enlarged board with numerous extra pieces that was popular at the court of Timur (about ad 1400).
The same game also included the dabbaba, whose name signifies a movable siege-engine. Pieces with the single and double diagonal steps, the fers and alfil, existed in mediaeval chess, as described in the Early History section. Their names derive from the Persian 'farzin' meaning counsellor and 'pil' meaning elephant. A piece with the knight move has existed in all forms of chess, and has always been named for the horse or a horseman.
The camel and giraffe (the latter with added power to move further after its leap) were pieces in Timur's Great Chess.
[This result was given, in slightly different form, by Edouard Lucas in his Recreations Mathematiques (vol.IV, 1894, p.130).]
In the case of {r,s} skew leapers, r < s, eight different mobility patterns can occur on square boards according to the size of the board n×n and the relative proportions of the numbers r, s and n. The case (b) s < n < 2s divides into five separate cases according to the value of r: (b1) s < n < 2r, (b2) n = 2r, (b3) 2r < n < r + s, (b4) n = r + s, (b5) r + s < n < 2s.
Cases (a), (b4), (c), (d) occur for all skew leapers. The other four cases only occur under special conditions: (b1) s < 2r – 1, (b2) s < 2r, (b3) r + 1 < s < 2r, (b5) r + 1 < s.
The mobility pattern of an {r,s} skew leaper on a square board of edge r+s has two moves available at every cell except the central ones, this means the moves must form closed circuits. The {1,2k}-movers form a single circuit. [I first gave the above results on mobility patterns in Chessics #24, 1985, p.88.]
Puzzle 1: What is the smallest board on which an {r,s} leaper can move from every cell?
Puzzle 2: On what board does the mobility of a {0,s} leaper equal 3?
Puzzle 3: Which is the smallest simple skew leaper that exhibits all eight mobility patterns?
Puzzle 4: On what size of board do the moves of an {r,s} leaper form circuits?
Puzzle 5: Which simple leapers on boards answering puzzle 4 form a single circuit?
A compound leaper is a piece that can make two or more different, non-null, patterns of leap. The simplest and best known of course is the king. Names for all combinations using coordinates less than 3 are:
{0,1} + {1,1} = king,
{0,1}+{0,2} = wazaba, {1,1}+{0,2} = diamond,
{0,1}+(1,2} = emperor, {1,1}+{1,2} = prince, {0,2}+{1,2} = templar,
{0,1}+{2,2} = caliph, {1,1}+{2,2} = ferfil, {0,2}+{2,2} = alibaba, {1,2}+{2,2} = hospitaler.
Some other combinations named by chess problemists are:
{0,2}+{1,2}+{2,2} = squirrel, {1,2}+{1,3} = gnu, {1,3}+{2,3} = bison.
An important group of longer-leaping compounds are the fixed-distance leapers. As longer leaps are considered cases arise in which different patterns of move have the same length. The first such cases, and the only ones that can move on the 8×8 board, are the {0,5}+{3,4} = fiveleaper, and the {5,5}+{1,7} = root-fifty-leaper.
T. R. Dawson gave an analysis of multi-pattern fixed-distance leapers on larger boards in Chess Amateur August 1925:
Double-pattern: {4,7}+{1,8} = root-65L, {6,7}+{2,9} = root-85L, {6,8}+{0,10} = 10L,
{5,10}+{2,11} = root-125L, {7,9}+{3,11} = root-130L, {8,9}+{1,12} = root-145L,
{5,12}+{0,13} = 13L, {7,11}+{1,13} = root-170L, {8,11}+{4,13} = root-185L,
{10,10}+{2,14} = double-root-50L, {6,13}+{3,14} = root-205L, {10,11}+{5,14} = root221L,
{9,12}+{0,15} = 15L, {9,13}+{5,15} = root-250L, {8,14}+{2,16} = double-root-65L,
{11,12}+{3,16} = root-265L, {8,15}+{0,17} = 17L, {11,13}+{1,17} = root-290L,
{7,16}+{4,17} = root-305L.
Triple-pattern: {10,15}+{6,17}+{1,16} = root-325L, {15,20}+{7,24}+{0,25} = 25L
Quadruple-pattern: {23,24}+{12,31}+{9,32}+{4,33} = root-1105L
Quintuple-pattern: {39,52}+{33,56}+{25,60}+{16,63}+{0,65} = 65L.
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Any sequence of moves constitutes a journey. The resulting change in position of the moved man is always equivalent to a single leap; we thus speak of a (p,q)-journey if the final cell reached is a (p,q) move from the initial cell. A journey is either open (ending at a different cell from that where it started) or closed (ending at the initial cell). A closed jourey is a (0,0)-journey.
Reordering Principle. THEOREM: On a sufficiently large board a series of moves will reach the same destination, from a given initial cell, regardless of the order in which the moves are made. Proof: Any move in the journey takes the piece a given distance horizontally and a given distance vertically (these distances may be positive, negative or zero). The sums of the horizontal and of the vertical moves give the coordinates of the end point relative to the start point. A sum of distances is independent of the order in which they are added.
Number of Journeys. Consider a (p,q) journey of j = a + b + c + d + a' + b' + c' + d' moves in the eight directions of an {r,s} skew-mover. The number of ways these moves can be reordered is, by well known algebra, elegantly expressed as j!/(a!b!c!...). For example, diagram (A) illustrates the 5!/2!2! = 30 knight journeys formed of two moves in the (1,2) direction, two in the (2,1) direction and one in the (2,–1) direction from a2 to i7, that is an (8,5)-journey of 5 moves. The same diagram also illustrates the 5! = 120 journeys formed of five moves, one in each of the directions (1,2), (2,1), (2,–1), (–1,–2), (–2,–1), from d5 to f4, but since two pairs of these moves are in opposite directions some of these journeys involve switchbacks and circuits.
The interesting patterns formed by the shortest journeys of a given leaper between two given cells were first pointed out to me by C.M.B.Tylor [Chessics vol.2 nr.19 p.30 1984]. Diagram (B) shows the 12 knight's paths of three moves between two adjacent cells, forming a pattern of two interlocking cubes. Diagram (C) shows the 24 knight's paths of four moves to make a (0,6) journey, forming a 'hypercube' pattern (the 4-dimensional equivalent of the cube).
Recreation: Try constructing all the shortest knight journeys between two arbitrarily chosen cells to see the pattern that appears. Draw four-move (0,10)-journeys by zebra and giraffe for comparison.
The Journey Equations. Adding horizontal and vertical components we get the journey equations:
(1a) p = [(a – a') + (d – d')]r + [(b – b') + (c – c')]s
(1b) q = [(a – a') + (d' – d)]s + [(b – b') + (c' – c)]r.
Substituting A = a – a', B = b – b', C = c – c', D = d – d' the journey equations take the simpler form:
(2a) p = (A + D)r + (B + C)s and (2b) q = (A – D)s + (B – C)r.
We will make much use of these equations subsequently.
Simplified Journeys. In a journey involving moves in opposite directions we can re-order the moves to bring the opposing moves together to cancel each other out. The journey of j moves is reduced by this method to a journey of i = |A| + |B| + |C| + |D| moves in four or fewer of the eight directions. A series of moves all in the same direction is a ride. Thus by reordering we reduce any journey to 4 or fewer rides in different lines.
The four rides can be taken in 4! = 24 different orders, forming a hyper-parallelogram pattern of the type illustrated below. The diagrams simplify considerably if fewer than 4 directions are used, becoming a parallelepiped (3! = 6) when three are used, parallelogram (2! = 2) when two are used, line (1! = 1) when only one is used, and point (0! = 1) when all 4 rides are zero.
Closed Journeys. In the case of a closed journey p = q = 0, so the journey equations reduce to:
0 = (A + D)r + (B + C)s and 0 = (A – D)s + (B – C)r,
from which, since s is not 0, we get
r/s = –(B + C)/(A + D) = –(A – D)/(B – C)
and multiplying through by –(A + D)(B –C) gives us
(B + C)(B – C) = (A + D)(A – D)
which is the same as B^2 – C^2 = A^2 – D^2.
We can now substitute the absolute values since the squares of negative numbers are positive, and so finally
|A|^2 + |C|^2 = |B|^2 + |D|^2.
Geometrically this relation means that the four rides (when none is zero) form two right angled triangles with a common longest side, A being perpendicular to C and B to D. This is obviously possible whenever A = B and C = D or when A = D and B = C in which cases the quadrilateral is a kite when arranged so that no two lines cross, or a bow-tie when they cross. But there are also skew cases such as (A,C) = (5,5), (B,D) = (1,7). The cases with perimeters up to 30 units are: 7,1;5,5 (18); 8,1;7,4 (20); 9,2;7,6 (24); 11,2;10,5 (28); 11,3;9,7 (30); 12,1;9,8 (30).
When one of the rides reduces to zero the three rides form a right-angled triangle. For an {r,s}-mover with r < s the sides of this triangle must be in the ratios s^2 – r^2 : 2rs : r^2 + s^2.
In the case of the knight we get the famous 3 : 4 : 5 triangle.
We can use the above results to provide an alternative proof of the following:
THEOREM: Any closed journey by an {r,s} skew leaper consists of an even number of moves.
Proof: For any journey we have i^2 = (|A| + |B| + |C| + |D|)^2 =
|A|^2 + |B|^2 + |C|^2 + |D|^2 + 2(|A||B| + |A||C| + |A||D| + |B||C| + |B||D| + |C||D|).
Therefore in the case of a circuit, since |A|^2 + |C|^2 = |B|^2 + |D|^2,
i^2 is even, and hence i also is even (since the square of an odd number would be odd). In other words we have proved that a circuit of an odd number of moves by a single-pattern leaper is impossible.
Compact Journeys. H.J.R.Murray (1942) considered the problem of arranging the moves in these simplified journeys (which, following Jaenisch (1862), he called 'irreducible chains') so as to pack them within the smallest possible board. The following are his results for the 2,2,1,1 and 3,3,1,1 and 5,4,3 cases, and my own best result for the 7,5,5,1 case.
Puzzle 7: Fit the 8,1;7,4 case onto a smallest possible board.
Shown is a ride pattern for all possible ways of making a closed journey of type 7,1;5,5.
Two-Move Journeys and Angles. Two-move journeys of a skew mover are of seven types when classified in terms of the equivalent {p,q} move or by the angle between the two lines. We will number the angles 0 to 6 so that, in the case of the knight, the number measures the angle to the nearest multiple of 30°. The angles can also be described as Null, Diagonal Acute, Lateral Acute, Right, Lateral Obtuse, Diagonal Obtuse and Straight.
There are always two choices for the right-angle move. When necessary we distinguish them as 3D and 3L (or just D and L) according to whether the second move crosses a diagonal or lateral from the starting point of the first move. The reverse journeys of 0, 1, 2, 4, 5, 6 have the same code, but each of the right-angle moves is the reverse of the other. Note that the distinction between 3D and 3L is not that between moves to right or left (clockwise or anticlockwise) either type can occur in either direction.
The two acute angles are a = 2 tan–1 [(s – r)/(s + r)] and b = 2 tan–1 (r/s), related by a + b = 90°. These angles can never be equal, since this would require r/s = (s – r)/(s + r) which implies (r + s)/s = Ö2, but the square root of two cannot be expressed as a ratio of whole numbers. The angle b is smallest if r/s < Ö2, but a is smallest if r/s > Ö2. The angles a and b are the acute angles of the right-angled triangle with sides in the proportions
(s^2 – r^2) : 2rs : (s^2 + r^2).
In the case of the knight this is the 3 : 4 : 5 triangle. Murray (1942) gives the values of these angles for the knight in degrees, minutes and seconds as follows: a = 36° 52' 11.4" and b = 53° 7' 48.6".
Puzzle 8: For which leaper on the 8×8 board are a and b most nearly equal?
Puzzle 9: What leapers other than {kr,ks} can make the same angles as {r,s}?
Angles in Closed Tours. Some general propositions concerning the numbers of the different angles in closed knight's tours were proved by Fred Schuh Wonderlijke Problemen (1943) [translated as The Master Book of Mathematical Recreations (1968)] and these are generalised here to closed journeys by free-leapers of all types. The number of angles of types 1, 3, 4, say, in a path is denoted by N(1, 3, 4).
(1) The total number of angles in a closed journey, including the angle where the last and first moves meet, is N(1, 2, 3, 4, 5, 6) = N(1) + N(2) + N(3) + N(4) + N(5) + N(6) which equals the total number of moves and is even, as we have proved earlier.
(2) The move pairs with angles 1, 3 and 5 move the leaper an odd number of ranks and files, since the numbers involved are s – r and s + r which are odd for free-leapers. It follows that N(1, 3, 5) is even in any closed journey (since the total number of ranks or files moved is even if the piece has to get back to its original rank and file — the same number out and back).
(3) Move-pairs with angles 2, 4 and 6 move the leaper an even number of ranks and files. From the previous two results we must have N(2, 4, 6) even also in a closed path.
(4) Move-pairs with angles 1, 2, 4 and 5 rotate the leaper's move by angles ±a, ±(90°–a), ±(90°+a) and ±(180°–a) clockwise or anticlockwise, and a and 90° are incommensurable (i.e. no whole number multiple of a is a whole number multiple of 90°).
The total angle moved through must be a multiple of 360°, since the angle between the last and first moves reorientates the piece to face in the same direction as it started out. Therefore N(1, 2, 4, 5) is even in a closed path, because positive and negative alphas must balance out.
(5) Combining results (4) and (1) tells us that N(3, 6) is even.
(6) Saying 'the sum of two quantities is even' can alternatively be expressed by saying that 'two quantities are of the same parity', that is, if one is odd the other is odd, and if one is even the other is even. Thus the above results can be summarised by saying that in a closed path the numbers of right (3), straight (6), diagonal (1 or 5) and lateral (2 or 4) angles are always of the same parity. In a symmetric path they are obviously all even.
Three-Move Journeys. Three-move journeys without switchbacks, considered directionally, are of 7×7 = 49 types, since at the second and third moves there is in each case a choice of 7 directions for the next move. We can indicate them by ordered pairs of the angle-codes for two-move journeys. There are 7 three-move paths that are of the same type when traversed in the opposite direction (11, 22, DL, LD, 44, 55, 66). There are 20 not using right angles that are of type uv reversing to vu (12/21, 14/41, 15/51, 16/61, 24/42, 25/52, 26/62, 45/54, 46/64, 56/65). Then there are 20 using one right angle, the reverse of uD being Lu and of uL being Du (u not D or L). There remain the pair of cases DD and LL, each the reverse of the other. The rule is to code the reverse journey is: reverse the sequence and replace D by L and L by D.
Puzzle 10: Why are DL and LD of the same length but in different directions?
Geometrically there are 7(7+1)/2 = 28 different 3-move paths: 11, 12/21, 1D/L1, 1L/D1, 14/41, 15/51, 16/61, 22, 2D/L2, 2L/D2, 24/42, 25/52, 26/62, LL/DD, DL, LD, L4/4D, 4L/D4, L5/5D, 5L/D5, L6/6D, 6L/D6, 44, 45/54, 46/64, 55, 56/65, 66.
There are only 12 types of triple leap when classified in terms of equivalent {p,q} move: 11, 16, = {|2r–s|, |2s–r|}, 12, 1D, 2L = {r, |2r–s|}, 1L, 14, L4 = {s, |2s–r|}, 15, 24, DD = {r,s}, 22, 26 = {3r,s}, 2D, 25, D5 = {2r+s,r}, DL, L6 = {|2s–r|, 2r+s}, LD, D6 = {2s+r, |2r–s|}, 34, L5, 45 = {s,2s+r}, 44, 46 = {r,3s}, 55, 56 ={2r+s, 2s+r}, 66 = {3r,3s}.
For the particular cases of knight and camel the types of three-move journeys reduce to fewer. For the knight {1,2} we have {s, 3r} = {2, 3} = {r, 2s–r}. For the camel {1, 3} we have {s–2r, 2s–r} = {1, 5}= {r, 2r+s}.
These results provide another proof that a triangle of moves by an {r,s} skew leaper is impossible. Since neither r nor s is zero the only case where {p,q} might be {0,0} is where subtractions occur in both coordinates. The only case is {|2r–s|, |2s–r|} which requires 2r = s and 2s = r which implies r = s = 0.
Four-Move Journeys. We can combine two 2-move journeys to form a 4-move circuit in three ways; by angle codes: 1515, 2424, DDDD (which is the same as LLLL). In coding a closed path of 2k moves all 2k angles are stated, the cyclic sequence being broken to give the smallest multi-digit number. Note that in a convex circuit the angle codes must add to 6k (counting D and L as 3).
In terms of sequences of angle codes four-move journeys without switchbacks number 7^3 = 343. This count includes 6 entries for the three four-move circuits (151, 515; 242, 424; DDD, LLL) and 6 entries for three skew-symmetric paths, having an axis of symmetry in an {r,s} direction and formed of right angles and straights (DLD/LDL; D6D/L6L; 6D6/6L6). The remaining 331 codes represent 150 asymmetric and 31 symmetric open paths (2×150 + 31 = 331). These symmetric paths of course give the same code when traversed in either direction.
Six-Move Circuits. By combining two three-move paths between the same cells we can form six-move circuits. The following two diagrams (using Giraffe {1,4} leaps as an example) show the 13 different circuits that are always possible with any {r,s} skew leaper. The 'open book' pattern gives 9 cases (AA', AB, AB', AC, AC', BB', BC, BC', CC' or in angle codes: 12DD25, 1241DD, 15L42D, 1D42L5, 14LL45, 156156, 245LL5, 246246, DD6DD6) while the 'cube' pattern gives 4 cases (ABC, ABD, ACD, BCD, where for example ABD denotes the hexagon ABD'A'B'D'; in angle codes: 12D12D, 1D41D4, 25D25D, D54D54).
In the case of knight {1,2} and camel {1,3} however, there are 12 further circuits possible in addition to the above 13, making 25 in all. [A note in Fairy Chess Review (November 1949 p.68) states this total was reported "long ago in Chess Amateur", but I have not been able to trace the exact reference.]
The reason for the extra circuits being possible in the case of the knight is that certain of the three-move paths, namely 11, 12, 1D, 16, D2, LD, 36, in which the end-points are separated by |2r–s| ranks or files in the general case are in the case of the knight on the same rank or file, since for the knight 2r = s. Similarly in the case of the camel the end-points of the three-move paths 12, 1D, 22, D2, DL, L6, 26, lie on a diagonal. Any knight path can in fact be converted into a camel path by increasing all lengths by a factor of root 2 and rotating through 45 degrees. This correspondence between knight and camel was first pointed out by T.R.Dawson in the Problemist Fairy Chess Supplement, June 1933, and has been independently rediscovered by several other workers since. The following are the twelve extra circuits (in angle code: 112112, 121D2L, 122214. 141626, 222L4D, 2LD2LD, 26L4D6, 11261D, 122L1D, 1262L4, 1D62LD, 1D61D6):
The above results account to a considerable extent for the greater touring ability of the knight as compared to other leapers and single it out for special study.
Eight-Move Symmetric Circuits. The symmetric 8-move circuits were enumerated by T.R.Dawson, finding 106, and he illustrated 100 of them in his series of knights tours with square numbers in symmetric closed knight chains (these are all diagrammed in my booklet on Figured Tours). The following are the results of a check on this enumeraton using our angle-coding method. The enumeration is quite tricky, there being several special cases easily overlooked. In particular there are three showing skew symmetry, which we show first (1 with octonary symmetry, 1 with biaxial symmetry and 1 with a single axis). By joining together two equal symmetric four-move paths we can form 10 circuits with biaxial or higher symmetry (4 octonary, 3 with lateral axes and 3 with diagonal axes).
By joining together two different symmetric four-move paths we can, for the general {r,s} skew leaper, form 36 symmetric circuits with a single axis of symmetry. There are 18 with lateral and 18 with diagonal symmetry, occurring in modally related pairs (10 without self-intersection); that is lateral angles in one correspond to diagonal angles in the other. There are also four special cases of axial symmetry formed by reflecting an asymmetric four-move path in the perpendicular bisector of the line joining its end-points (2 with lateral and 2 with diagonal axis).
In the special cases of knight and camel we have to add a further 16 circuits formed by the first method, and 4 formed by the second method, all with diagonal axis for the knight but lateral axis for the camel. These connections are not possible in the general case since the end-points of the two component paths are at different distances apart in the first method, and are not in the same lateral or diagonal in the second method.
We now come to the cases with pure rotational symmetry. By joining an asymmetric four-move path of an {r,s} skew leaper with a 180° rotated copy of itself we can form a symmetric circuit with pure rotatory symmetry in 29 cases, as shown below, classified into 3 self-modal forms (3564, 1623, 1254) and 13 modally related pairs. The 29 can also be classified into 12 centred on the corner of a cell, and 17 on the centre of a cell. In the case of the knight 8 of the 29 circuits self-intersect, but this is not necessarily the case for other pieces: while the circuit 2356 is non-intersecting for the knight or zebra, its modal twin 1346 intersects, but for the giraffe, the first is intersecting and the second not.
It is very easy to miss four further circuits with pure rotational symmetry that are possible in the case of knight and camel. These have Bergholtian symmetry, in which the path crosses itself at the centre. These cannot be split into two equal four-move paths, but can be analysed into two equal three-move paths whose ends in the case of the knight are separated by {0,1} or {0,2} moves thus permitting them to be joined to the X of knight moves (in the case of the camel moves the ends of the X are separated by {1,1} and {2,2} moves). Alternatively they can be analysed into two three-move paths with rotary symmetry which cross at their mid-points and whose ends can be joined.
For the fers {1,1} we must have p+q even, since it is confined to squares of one colour. Thus, assuming p £ q the minimum number of moves required is i = q, composed of (q–p)/2 in one direction and p + (q–p)/2 = (p+q)/2 at right angles. Illustrated are the q!/[(p+q)/2]![(q–p)/2]! = 4 routes of a fers in making a {2,4} journey. The more general {r,r}-mover requires i = q/r moves.
Example of restriction to a board with fixed edges: the knight takes two moves minimum to make a diagonal step (e.g. a2-c3-b1) but in the corner of most rectangular boards requires four moves to get from a1 to b2 (e.g. a1-b3-d2-c4-b2) and on the 3×3 board cannot make the journey at all.
Shortest Leaper Paths. The case of a closed path, when {p,q} = {0,0} was considered in the last section. The shortest path is the null path, which means the piece just remains where it is. If it has to move a positive distance then the shortest journey is a switchback (one move out and back again). If it must not retrace any section of its path the shortest journey is a circuit of four moves, of which there are three geometrically different types (squares, diamonds and lozenges).
If we must not make two moves in opposite directions the shortest path is of six moves: four rides of lengths 1, 1, 2, 2 in some sequence, forming a kite or bow shape. If we require a journey in the minimum number of rides without switchbacks, the answer is the 3, 4, 5 triangle of 12 moves. [??? is this true generally or only for knight???]
The problem in the above form was posed by A.I.Houston in the 1970s and he solved it for {r,s} = {z,z+1}, a case that includes knight, zebra and antelope. We see that (z+1)–z = 1, so x = –1 and y = 1. Trying m = 1 and n = 1 we calculate from (5): A = 1, B = 0, C = –z, D = z, so i = 2z+1.
Another family of cases is {r,s} = {1,2z}, which includes knight and giraffe. Here we can use 1+0(2z) = 1, which gives x = 1, y = 0. Taking m = 1 and n = 0 we find from (5): A = z, B = 0, C = –1, D = z and so again i = 2z+1.
These two cases, illustrated below, conform to the rule i = r+s. In general however r+s £ i.
Puzzle 8: For which leaper on the 8×8 board are a and b most nearly equal? Answer: the {2,5} leaper is slightly closer than the {3,7} leaper, since 2/5 = 0.4 and 3/7 = 0.428571. The first is out by 0.014214 and the second by 0.014357.
Puzzle 9: What leapers other than {kr,ks} can make the same angles as {r,s}? Answer the {s–r, s+r} leaper. This result is due to O. E. Vinje, (Fairy Chess Review December 1940, problem 4656). In particular the camel {1,3} has the same angles between its moves as the knight {1,2}.
Puzzle 10: Why are DL and LD of the same length but in different directions? Answer; the moves DL = {|2r–s|, 2s+r} and LD = {2r+s, 2s–r} are always different, because equality would entail either 2s+r = 2r+s or 2s+r = 2s–r, and these conditions imply r=s or r=0 respectively, contradicting the assumption that the leaper is skew.
The fact that they are of the same length is obvious from geometrical congruence, but this can be confirmed by algebra as follows: (2s–r)^2 + (2r+s)^2 = 5s^2 + 5r^2 = (2r–s)^2 + (2s+r)^2. The length of the triple leap is root 5 times the length of the {r,s} move.